3.15.22 \(\int \frac {-10 x^3 \log (x) \log ^2(3 x)+e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} (x^2 \log (x) \log ^2(3 x)+e^{\frac {25}{\log (3 x)}} (-25 \log (x)-\log (x) \log ^2(3 x)))+(-e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} x \log ^2(3 x)+5 x^3 \log ^2(3 x)) \log (\frac {1}{5} (e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}}-5 x^2))}{e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} x^2 \log ^2(x) \log ^2(3 x)-5 x^4 \log ^2(x) \log ^2(3 x)} \, dx\)

Optimal. Leaf size=34 \[ \frac {\log \left (\frac {1}{5} e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}-x^2\right )}{\log (x)} \]

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Rubi [F]  time = 32.28, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-10 x^3 \log (x) \log ^2(3 x)+e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} \left (x^2 \log (x) \log ^2(3 x)+e^{\frac {25}{\log (3 x)}} \left (-25 \log (x)-\log (x) \log ^2(3 x)\right )\right )+\left (-e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} x \log ^2(3 x)+5 x^3 \log ^2(3 x)\right ) \log \left (\frac {1}{5} \left (e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}}-5 x^2\right )\right )}{e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} x^2 \log ^2(x) \log ^2(3 x)-5 x^4 \log ^2(x) \log ^2(3 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-10*x^3*Log[x]*Log[3*x]^2 + E^((E^(25/Log[3*x]) + x^2)/x)*(x^2*Log[x]*Log[3*x]^2 + E^(25/Log[3*x])*(-25*L
og[x] - Log[x]*Log[3*x]^2)) + (-(E^((E^(25/Log[3*x]) + x^2)/x)*x*Log[3*x]^2) + 5*x^3*Log[3*x]^2)*Log[(E^((E^(2
5/Log[3*x]) + x^2)/x) - 5*x^2)/5])/(E^((E^(25/Log[3*x]) + x^2)/x)*x^2*Log[x]^2*Log[3*x]^2 - 5*x^4*Log[x]^2*Log
[3*x]^2),x]

[Out]

LogIntegral[x] + Defer[Int][E^(E^(25/Log[3*x])/x + x + 25/Log[3*x])/(x^2*(-E^(E^(25/(Log[3] + Log[x]))/x + x)
+ 5*x^2)*Log[x]), x] + 10*Defer[Int][x/((-E^(E^(25/Log[3*x])/x + x) + 5*x^2)*Log[x]), x] - 5*Defer[Int][x^2/((
-E^(E^(25/Log[3*x])/x + x) + 5*x^2)*Log[x]), x] + 25*Defer[Int][E^(E^(25/Log[3*x])/x + x + 25/Log[3*x])/(x^2*(
-E^(E^(25/Log[3*x])/x + x) + 5*x^2)*Log[x]*Log[3*x]^2), x] - Defer[Int][Log[(E^(E^(25/Log[3*x])/x + x) - 5*x^2
)/5]/(x*Log[x]^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-10 x^3 \log (x) \log ^2(3 x)+e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} \left (x^2 \log (x) \log ^2(3 x)+e^{\frac {25}{\log (3 x)}} \left (-25 \log (x)-\log (x) \log ^2(3 x)\right )\right )+\left (-e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} x \log ^2(3 x)+5 x^3 \log ^2(3 x)\right ) \log \left (\frac {1}{5} \left (e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}}-5 x^2\right )\right )}{x^2 \left (e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}-5 x^2\right ) \log ^2(x) \log ^2(3 x)} \, dx\\ &=\int \left (-\frac {e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x+\frac {25}{\log (3 x)}} \left (25+\log ^2(3 x)\right )}{x^2 \left (e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}-5 x^2\right ) \log (x) \log ^2(3 x)}+\frac {-e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x} x \log (x)+10 x^2 \log (x)+e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x} \log \left (\frac {1}{5} \left (e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}-5 x^2\right )\right )-5 x^2 \log \left (\frac {1}{5} \left (e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}-5 x^2\right )\right )}{x \left (-e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}+5 x^2\right ) \log ^2(x)}\right ) \, dx\\ &=-\int \frac {e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x+\frac {25}{\log (3 x)}} \left (25+\log ^2(3 x)\right )}{x^2 \left (e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}-5 x^2\right ) \log (x) \log ^2(3 x)} \, dx+\int \frac {-e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x} x \log (x)+10 x^2 \log (x)+e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x} \log \left (\frac {1}{5} \left (e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}-5 x^2\right )\right )-5 x^2 \log \left (\frac {1}{5} \left (e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}-5 x^2\right )\right )}{x \left (-e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}+5 x^2\right ) \log ^2(x)} \, dx\\ &=-\int \left (-\frac {e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x+\frac {25}{\log (3 x)}}}{x^2 \left (-e^{\frac {e^{\frac {25}{\log (3)+\log (x)}}}{x}+x}+5 x^2\right ) \log (x)}-\frac {25 e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x+\frac {25}{\log (3 x)}}}{x^2 \left (-e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}+5 x^2\right ) \log (x) \log ^2(3 x)}\right ) \, dx+\int \left (-\frac {5 (-2+x) x}{\left (-e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}+5 x^2\right ) \log (x)}+\frac {x \log (x)-\log \left (\frac {1}{5} \left (e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}-5 x^2\right )\right )}{x \log ^2(x)}\right ) \, dx\\ &=-\left (5 \int \frac {(-2+x) x}{\left (-e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}+5 x^2\right ) \log (x)} \, dx\right )+25 \int \frac {e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x+\frac {25}{\log (3 x)}}}{x^2 \left (-e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}+5 x^2\right ) \log (x) \log ^2(3 x)} \, dx+\int \frac {e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x+\frac {25}{\log (3 x)}}}{x^2 \left (-e^{\frac {e^{\frac {25}{\log (3)+\log (x)}}}{x}+x}+5 x^2\right ) \log (x)} \, dx+\int \frac {x \log (x)-\log \left (\frac {1}{5} \left (e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}-5 x^2\right )\right )}{x \log ^2(x)} \, dx\\ &=-\left (5 \int \left (-\frac {2 x}{\left (-e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}+5 x^2\right ) \log (x)}+\frac {x^2}{\left (-e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}+5 x^2\right ) \log (x)}\right ) \, dx\right )+25 \int \frac {e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x+\frac {25}{\log (3 x)}}}{x^2 \left (-e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}+5 x^2\right ) \log (x) \log ^2(3 x)} \, dx+\int \frac {e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x+\frac {25}{\log (3 x)}}}{x^2 \left (-e^{\frac {e^{\frac {25}{\log (3)+\log (x)}}}{x}+x}+5 x^2\right ) \log (x)} \, dx+\int \left (\frac {1}{\log (x)}-\frac {\log \left (\frac {1}{5} \left (e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}-5 x^2\right )\right )}{x \log ^2(x)}\right ) \, dx\\ &=-\left (5 \int \frac {x^2}{\left (-e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}+5 x^2\right ) \log (x)} \, dx\right )+10 \int \frac {x}{\left (-e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}+5 x^2\right ) \log (x)} \, dx+25 \int \frac {e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x+\frac {25}{\log (3 x)}}}{x^2 \left (-e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}+5 x^2\right ) \log (x) \log ^2(3 x)} \, dx+\int \frac {1}{\log (x)} \, dx+\int \frac {e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x+\frac {25}{\log (3 x)}}}{x^2 \left (-e^{\frac {e^{\frac {25}{\log (3)+\log (x)}}}{x}+x}+5 x^2\right ) \log (x)} \, dx-\int \frac {\log \left (\frac {1}{5} \left (e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}-5 x^2\right )\right )}{x \log ^2(x)} \, dx\\ &=\text {li}(x)-5 \int \frac {x^2}{\left (-e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}+5 x^2\right ) \log (x)} \, dx+10 \int \frac {x}{\left (-e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}+5 x^2\right ) \log (x)} \, dx+25 \int \frac {e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x+\frac {25}{\log (3 x)}}}{x^2 \left (-e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}+5 x^2\right ) \log (x) \log ^2(3 x)} \, dx+\int \frac {e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x+\frac {25}{\log (3 x)}}}{x^2 \left (-e^{\frac {e^{\frac {25}{\log (3)+\log (x)}}}{x}+x}+5 x^2\right ) \log (x)} \, dx-\int \frac {\log \left (\frac {1}{5} \left (e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}-5 x^2\right )\right )}{x \log ^2(x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.76, size = 34, normalized size = 1.00 \begin {gather*} \frac {\log \left (\frac {1}{5} \left (e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}-5 x^2\right )\right )}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10*x^3*Log[x]*Log[3*x]^2 + E^((E^(25/Log[3*x]) + x^2)/x)*(x^2*Log[x]*Log[3*x]^2 + E^(25/Log[3*x])*
(-25*Log[x] - Log[x]*Log[3*x]^2)) + (-(E^((E^(25/Log[3*x]) + x^2)/x)*x*Log[3*x]^2) + 5*x^3*Log[3*x]^2)*Log[(E^
((E^(25/Log[3*x]) + x^2)/x) - 5*x^2)/5])/(E^((E^(25/Log[3*x]) + x^2)/x)*x^2*Log[x]^2*Log[3*x]^2 - 5*x^4*Log[x]
^2*Log[3*x]^2),x]

[Out]

Log[(E^(E^(25/Log[3*x])/x + x) - 5*x^2)/5]/Log[x]

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fricas [A]  time = 1.74, size = 33, normalized size = 0.97 \begin {gather*} \frac {\log \left (-x^{2} + \frac {1}{5} \, e^{\left (\frac {x^{2} + e^{\left (\frac {25}{\log \relax (3) + \log \relax (x)}\right )}}{x}\right )}\right )}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*log(3*x)^2*exp((exp(25/log(3*x))+x^2)/x)+5*x^3*log(3*x)^2)*log(1/5*exp((exp(25/log(3*x))+x^2)/x
)-x^2)+((-log(x)*log(3*x)^2-25*log(x))*exp(25/log(3*x))+x^2*log(x)*log(3*x)^2)*exp((exp(25/log(3*x))+x^2)/x)-1
0*x^3*log(x)*log(3*x)^2)/(x^2*log(x)^2*log(3*x)^2*exp((exp(25/log(3*x))+x^2)/x)-5*x^4*log(x)^2*log(3*x)^2),x,
algorithm="fricas")

[Out]

log(-x^2 + 1/5*e^((x^2 + e^(25/(log(3) + log(x))))/x))/log(x)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*log(3*x)^2*exp((exp(25/log(3*x))+x^2)/x)+5*x^3*log(3*x)^2)*log(1/5*exp((exp(25/log(3*x))+x^2)/x
)-x^2)+((-log(x)*log(3*x)^2-25*log(x))*exp(25/log(3*x))+x^2*log(x)*log(3*x)^2)*exp((exp(25/log(3*x))+x^2)/x)-1
0*x^3*log(x)*log(3*x)^2)/(x^2*log(x)^2*log(3*x)^2*exp((exp(25/log(3*x))+x^2)/x)-5*x^4*log(x)^2*log(3*x)^2),x,
algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.23, size = 34, normalized size = 1.00

method	result	size
risch	\(\frac {\ln \left (\frac {{\mathrm e}^{\frac {{\mathrm e}^{\frac {25}{\ln \relax (x )+\ln \relax (3)}}+x^{2}}{x}}}{5}-x^{2}\right )}{\ln \relax (x )}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x*ln(3*x)^2*exp((exp(25/ln(3*x))+x^2)/x)+5*x^3*ln(3*x)^2)*ln(1/5*exp((exp(25/ln(3*x))+x^2)/x)-x^2)+((-l
n(x)*ln(3*x)^2-25*ln(x))*exp(25/ln(3*x))+x^2*ln(x)*ln(3*x)^2)*exp((exp(25/ln(3*x))+x^2)/x)-10*x^3*ln(x)*ln(3*x
)^2)/(x^2*ln(x)^2*ln(3*x)^2*exp((exp(25/ln(3*x))+x^2)/x)-5*x^4*ln(x)^2*ln(3*x)^2),x,method=_RETURNVERBOSE)

[Out]

1/ln(x)*ln(1/5*exp((exp(25/(ln(x)+ln(3)))+x^2)/x)-x^2)

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maxima [A]  time = 1.01, size = 35, normalized size = 1.03 \begin {gather*} -\frac {\log \relax (5) - \log \left (-5 \, x^{2} + e^{\left (x + \frac {e^{\left (\frac {25}{\log \relax (3) + \log \relax (x)}\right )}}{x}\right )}\right )}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*log(3*x)^2*exp((exp(25/log(3*x))+x^2)/x)+5*x^3*log(3*x)^2)*log(1/5*exp((exp(25/log(3*x))+x^2)/x
)-x^2)+((-log(x)*log(3*x)^2-25*log(x))*exp(25/log(3*x))+x^2*log(x)*log(3*x)^2)*exp((exp(25/log(3*x))+x^2)/x)-1
0*x^3*log(x)*log(3*x)^2)/(x^2*log(x)^2*log(3*x)^2*exp((exp(25/log(3*x))+x^2)/x)-5*x^4*log(x)^2*log(3*x)^2),x,
algorithm="maxima")

[Out]

-(log(5) - log(-5*x^2 + e^(x + e^(25/(log(3) + log(x)))/x)))/log(x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^{\frac {25}{\ln \left (3\,x\right )}}+x^2}{x}}\,\left ({\mathrm {e}}^{\frac {25}{\ln \left (3\,x\right )}}\,\left (\ln \relax (x)\,{\ln \left (3\,x\right )}^2+25\,\ln \relax (x)\right )-x^2\,{\ln \left (3\,x\right )}^2\,\ln \relax (x)\right )-\ln \left (\frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^{\frac {25}{\ln \left (3\,x\right )}}+x^2}{x}}}{5}-x^2\right )\,\left (5\,x^3\,{\ln \left (3\,x\right )}^2-x\,{\ln \left (3\,x\right )}^2\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{\frac {25}{\ln \left (3\,x\right )}}+x^2}{x}}\right )+10\,x^3\,{\ln \left (3\,x\right )}^2\,\ln \relax (x)}{5\,x^4\,{\ln \left (3\,x\right )}^2\,{\ln \relax (x)}^2-x^2\,{\ln \left (3\,x\right )}^2\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{\frac {25}{\ln \left (3\,x\right )}}+x^2}{x}}\,{\ln \relax (x)}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((exp(25/log(3*x)) + x^2)/x)*(exp(25/log(3*x))*(25*log(x) + log(3*x)^2*log(x)) - x^2*log(3*x)^2*log(x)
) - log(exp((exp(25/log(3*x)) + x^2)/x)/5 - x^2)*(5*x^3*log(3*x)^2 - x*log(3*x)^2*exp((exp(25/log(3*x)) + x^2)
/x)) + 10*x^3*log(3*x)^2*log(x))/(5*x^4*log(3*x)^2*log(x)^2 - x^2*log(3*x)^2*exp((exp(25/log(3*x)) + x^2)/x)*l
og(x)^2),x)

[Out]

int((exp((exp(25/log(3*x)) + x^2)/x)*(exp(25/log(3*x))*(25*log(x) + log(3*x)^2*log(x)) - x^2*log(3*x)^2*log(x)
) - log(exp((exp(25/log(3*x)) + x^2)/x)/5 - x^2)*(5*x^3*log(3*x)^2 - x*log(3*x)^2*exp((exp(25/log(3*x)) + x^2)
/x)) + 10*x^3*log(3*x)^2*log(x))/(5*x^4*log(3*x)^2*log(x)^2 - x^2*log(3*x)^2*exp((exp(25/log(3*x)) + x^2)/x)*l
og(x)^2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*ln(3*x)**2*exp((exp(25/ln(3*x))+x**2)/x)+5*x**3*ln(3*x)**2)*ln(1/5*exp((exp(25/ln(3*x))+x**2)/x
)-x**2)+((-ln(x)*ln(3*x)**2-25*ln(x))*exp(25/ln(3*x))+x**2*ln(x)*ln(3*x)**2)*exp((exp(25/ln(3*x))+x**2)/x)-10*
x**3*ln(x)*ln(3*x)**2)/(x**2*ln(x)**2*ln(3*x)**2*exp((exp(25/ln(3*x))+x**2)/x)-5*x**4*ln(x)**2*ln(3*x)**2),x)

[Out]

Timed out

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